$$$2 \tan^{2}{\left(\theta \right)}$$$ 的積分

求$$$\int 2 \tan^{2}{\left(\theta \right)}\, d\theta$$$。

套用常數倍法則 $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$，使用 $$$c=2$$$ 與 $$$f{\left(\theta \right)} = \tan^{2}{\left(\theta \right)}$$$：

$${\color{red}{\int{2 \tan^{2}{\left(\theta \right)} d \theta}}} = {\color{red}{\left(2 \int{\tan^{2}{\left(\theta \right)} d \theta}\right)}}$$

令 $$$u=\tan{\left(\theta \right)}$$$。

則 $$$\theta=\operatorname{atan}{\left(u \right)}$$$ 與 $$$d\theta=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du = \frac{du}{u^{2} + 1}$$$（步驟見»）。

所以，

$$2 {\color{red}{\int{\tan^{2}{\left(\theta \right)} d \theta}}} = 2 {\color{red}{\int{\frac{u^{2}}{u^{2} + 1} d u}}}$$

重寫並拆分分式:

$$2 {\color{red}{\int{\frac{u^{2}}{u^{2} + 1} d u}}} = 2 {\color{red}{\int{\left(1 - \frac{1}{u^{2} + 1}\right)d u}}}$$

逐項積分：

$$2 {\color{red}{\int{\left(1 - \frac{1}{u^{2} + 1}\right)d u}}} = 2 {\color{red}{\left(\int{1 d u} - \int{\frac{1}{u^{2} + 1} d u}\right)}}$$

配合 $$$c=1$$$，應用常數法則 $$$\int c\, du = c u$$$：

$$- 2 \int{\frac{1}{u^{2} + 1} d u} + 2 {\color{red}{\int{1 d u}}} = - 2 \int{\frac{1}{u^{2} + 1} d u} + 2 {\color{red}{u}}$$

$$$\frac{1}{u^{2} + 1}$$$ 的積分是 $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$：

$$2 u - 2 {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}} = 2 u - 2 {\color{red}{\operatorname{atan}{\left(u \right)}}}$$

回顧一下 $$$u=\tan{\left(\theta \right)}$$$：

$$- 2 \operatorname{atan}{\left({\color{red}{u}} \right)} + 2 {\color{red}{u}} = - 2 \operatorname{atan}{\left({\color{red}{\tan{\left(\theta \right)}}} \right)} + 2 {\color{red}{\tan{\left(\theta \right)}}}$$

因此，

$$\int{2 \tan^{2}{\left(\theta \right)} d \theta} = 2 \tan{\left(\theta \right)} - 2 \operatorname{atan}{\left(\tan{\left(\theta \right)} \right)}$$

化簡：

$$\int{2 \tan^{2}{\left(\theta \right)} d \theta} = 2 \left(- \theta + \tan{\left(\theta \right)}\right)$$

加上積分常數：

$$\int{2 \tan^{2}{\left(\theta \right)} d \theta} = 2 \left(- \theta + \tan{\left(\theta \right)}\right)+C$$

$$$\int 2 \tan^{2}{\left(\theta \right)}\, d\theta = 2 \left(- \theta + \tan{\left(\theta \right)}\right) + C$$$A